[proofplan] The proof uses the degree-one [Hodge decomposition for compact Kähler manifolds](/theorems/8066). This decomposes $H^1(X;\mathbb C)$ as the direct sum of its $(1,0)$ and $(0,1)$ Hodge pieces. Complex conjugation interchanges these two summands, so they have the same complex dimension. Finally, the first Betti number is the complex dimension of $H^1(X;\mathbb C)$, giving the stated formula. [/proofplan]

[step:Decompose the first complex cohomology into Hodge summands]By [citetheorem:8066] applied in degree $1$, the complex cohomology group $H^1(X;\mathbb C)$ has the [Hodge decomposition](/theorems/2745) \begin{align*} H^1(X;\mathbb C)=H^{1,0}(X)\oplus H^{0,1}(X). \end{align*} The hypotheses of [citetheorem:8066] are satisfied because $X$ is compact Kähler by assumption.[/step]

[guided]We first use the part of the Kähler package that supplies the [Hodge decomposition](/theorems/3941). The theorem [citetheorem:8066] applies to compact Kähler manifolds, and $X$ satisfies exactly that hypothesis. Taking degree $k=1$ in its decomposition gives \begin{align*} H^1(X;\mathbb C)=\bigoplus_{p+q=1}H^{p,q}(X). \end{align*} The only pairs of nonnegative integers satisfying $p+q=1$ are $(p,q)=(1,0)$ and $(p,q)=(0,1)$. Therefore the decomposition becomes \begin{align*} H^1(X;\mathbb C)=H^{1,0}(X)\oplus H^{0,1}(X). \end{align*} This is the point where compactness and the Kähler condition are used: they are precisely the hypotheses that give this decomposition of cohomology into Hodge types.[/guided]

[step:Use Hodge symmetry to identify the dimensions of the two summands] By the Hodge-number symmetry [citetheorem:8081], applied with $(p,q)=(1,0)$, \begin{align*} h^{1,0}(X)=h^{0,1}(X). \end{align*} Equivalently, \begin{align*} \dim_{\mathbb C} H^{1,0}(X)=\dim_{\mathbb C} H^{0,1}(X). \end{align*} [/step]

[step:Compute the dimension of $H^1(X;\mathbb C)$] Since the decomposition above is a direct sum of complex vector spaces, finite-dimensionality and additivity of dimension give \begin{align*} \dim_{\mathbb C}H^1(X;\mathbb C)=\dim_{\mathbb C}H^{1,0}(X)+\dim_{\mathbb C}H^{0,1}(X). \end{align*} Using the equality of dimensions from the previous step, this becomes \begin{align*} \dim_{\mathbb C}H^1(X;\mathbb C)=2\dim_{\mathbb C}H^{1,0}(X). \end{align*} By the definition of the Hodge number $h^{1,0}(X)$, \begin{align*} \dim_{\mathbb C}H^1(X;\mathbb C)=2h^{1,0}(X). \end{align*} [/step]

[step:Identify the first Betti number and conclude evenness] The complexification map gives \begin{align*} H^1(X;\mathbb C)\cong H^1(X;\mathbb R)\otimes_{\mathbb R}\mathbb C, \end{align*} so \begin{align*} \dim_{\mathbb C}H^1(X;\mathbb C)=\dim_{\mathbb R}H^1(X;\mathbb R)=b_1(X). \end{align*} Combining this with the previous step yields \begin{align*} b_1(X)=2h^{1,0}(X). \end{align*} Since $h^{1,0}(X)$ is a nonnegative integer, $b_1(X)$ is twice an integer. Hence $b_1(X)$ is even. [/step]